Problem: Compute the value of $k$ such that the equation
\[\frac{x + 2}{kx - 1} = x\]has exactly one solution.
Answer: Assume $k \neq 0.$  Then
\[x + 2 = x(kx - 1) = kx^2 - x,\]so $kx^2 - 2x - 2 = 0.$  This quadratic has exactly one solution if its discriminant is 0, or $(-2)^2 - 4(k)(-2) = 4 + 8k = 0.$  Then $k = -\frac{1}{2}.$  But then
\[-\frac{1}{2} x^2 - 2x - 2 = 0,\]or $x^2 + 4x + 4 = (x + 2)^2 = 0,$ which means $x = -2,$ and
\[\frac{x + 2}{kx - 1} = \frac{x + 2}{-\frac{1}{2} x - 1}\]is not defined for $x = -2.$

So, we must have $k = 0.$  For $k = 0,$ the equation is
\[\frac{x + 2}{-1} = x,\]which yields $x = -1.$  Hence, $k = \boxed{0}$ is the value we seek.